# Newton's Second Law

> AP Physics 1 · Unit 2: Dynamics
> Source: https://www.owlsprep.com/study/ap-physics-1-u2-newton-s-second-law/

This module covers net force, vector form of Newton's second law, mass vs weight, accelerated motion on inclines, connected object systems, and problem-solving for single and multi-object systems for AP Physics 1.

**Prerequisites:** [Drawing free-body diagrams for rigid objects](https://www.owlsprep.com/study/ap-physics-1-u2-free-body-diagrams/); Resolving vectors into perpendicular components; Newton's first and third laws of motion

## Learning objectives

- State Newton's second law in vector form and relate net force, mass, and acceleration
- Distinguish between inertial mass and gravitational weight
- Solve 1D and 2D Newton's second law problems including inclined planes
- Apply the system approach to solve connected object acceleration problems
- Avoid common exam traps related to Newton's second law

## Core Definition of Newton's Second Law

Newton's Second Law is the core quantitative relationship between force, mass, and acceleration that underpins all of classical dynamics. It is the highest-weight topic in Unit 2 (Dynamics) for AP Physics 1, making up roughly 12-18% of the total exam score, appearing in both multiple-choice and free-response sections.

**Newton's Second Law** — The acceleration of an object is directly proportional to the net external force acting on it and inversely proportional to its inertial mass. The law holds independently for each vector component of force and acceleration.

*Notation:* \vec{F}_{\text{net}} = m\vec{a}

*Example:* Zero net force produces zero acceleration, which recovers Newton's first law as a special case.

> **warning**
>
> A common misconception is that Newton's second law relates net force to velocity. It only relates net force to acceleration: zero net force means zero acceleration, not zero speed.

**Check your understanding**

Test your understanding of variable relationships:

1. An experiment tests Newton's second law: a cart on a frictionless track is pulled by different hanging masses, with net force and acceleration measured for each trial. If acceleration is plotted on the y-axis and net force on the x-axis, what is the physical meaning of the slope?

   - A) The mass of the cart
   - B) The reciprocal of the mass of the cart
   - C) The weight of the cart
   - D) The acceleration due to gravity

   *Why:* Rearranging $F_{\text{net}} = ma$ gives $a = \left(\frac{1}{m}\right)F_{\text{net}}$, which matches the linear form $y = mx + b$, so the slope equals $1/m$.

## 1D Applications & Mass vs Weight

Because $\vec{F}_{\text{net}} = m\vec{a}$ is a vector equation, it holds independently for each perpendicular component of force and acceleration. The standard problem-solving workflow for all Newton's second law problems is:

1. Draw a complete free-body diagram for the object(s)
2. Choose a coordinate system aligned with the direction of acceleration
3. Resolve all forces into x and y components
4. Sum components to find net force in each direction
5. Set net force equal to $ma$ for each direction and solve for unknowns

**Mass vs Weight** — Mass is an invariant measure of an object's inertial resistance to acceleration (units: kg). Weight is the gravitational force acting on an object (units: N), near Earth's surface.

*Notation:* W = mg

**Worked example:** A 65 kg skydiver falls straight down, and experiences an upward drag force of 420 N. What is her acceleration?

1. Choose coordinate system with downward as the positive x direction (aligned with acceleration).
2. Identify all forces: downward weight $W = mg$, upward drag $F_d$.
3. Calculate net force:
4. $$F_{\text{net}} = W - F_d = (65)(9.8) - 420 = 637 - 420 = 217 \text{ N}$$
5. Rearrange Newton's second law to solve for acceleration:
6. $$a = \frac{F_{\text{net}}}{m} = \frac{217}{65} \approx 3.3 \text{ m/s}^2 \text{ downward}$$

> **tip**
>
> Always choose your coordinate system so that the direction of acceleration lies along one of the axes. This eliminates cross terms, saves time, and drastically reduces sign errors on the exam.

## Motion on Inclined Planes

Inclined plane problems are the most common 2-D Newton's second law problem on AP Physics 1, testing your ability to correctly decompose vectors and apply physical constraints to acceleration. The standard coordinate system for inclines aligns the x-axis parallel to the incline (direction of possible motion) and y-axis perpendicular to the incline.

Only weight is not aligned with the axes, so we decompose it into two components: parallel to the incline, $W_{\parallel} = mg \sin\theta$, and perpendicular to the incline, $W_{\perp} = mg \cos\theta$, where $\theta$ is the angle of the incline from the horizontal. For any fixed incline, an object cannot accelerate through the surface or jump off it unless explicitly stated, so $a_y = 0$, meaning net force perpendicular to the incline is always zero.

> **tip**
>
> To avoid swapping sine and cosine for weight components, check the edge case $\theta = 90^\circ$ (vertical incline = free fall). At this angle, $a = g$, which only works if the parallel component uses sine ($\sin90^\circ = 1$).

**Worked example:** A 12 kg box slides down a frictionless 25° incline. Find the magnitude of the box's acceleration.

1. Align x parallel to the incline (positive downward), y perpendicular (positive outward from the surface).
2. Decompose weight: $W_x = mg \sin25^\circ$, $W_y = -mg \cos25^\circ$. Normal force $N = +N$ in the y-direction, no other forces act on the box.
3. Apply Newton's second law to the y-direction first, where $a_y = 0$:
4. $$F_{\text{net},y} = N - mg \cos25^\circ = ma_y = 0 \implies N = mg \cos25^\circ$$
5. Apply Newton's second law to the x-direction:
6. $$F_{\text{net},x} = mg \sin25^\circ = ma_x$$
7. Mass cancels out, leaving:
8. $$a = g \sin25^\circ \approx 9.8 (0.42) \approx 4.1 \text{ m/s}^2$$

## Connected Objects & System Approach

When multiple objects are connected by a taut massless string over a massless, frictionless pulley, they move with the same magnitude of acceleration, so we can solve the system two ways: isolate each object and solve a system of equations, or treat the entire connected group as a single system.

For the system approach, internal forces (like tension in the connecting string between the objects) cancel out by Newton's third law, so we only need to include external forces when calculating net force. This drastically simplifies finding the acceleration of the system, but if we need to find tension itself, we still have to isolate one object to solve for it.

**Worked example:** A 3.0 kg block on a horizontal frictionless table is connected by a massless string over a massless pulley to a 1.5 kg hanging block. When released, what is the acceleration of the blocks?

1. Confirm the magnitude of acceleration is the same for both blocks, so the system approach is valid.
2. Mark the system boundary around both blocks: tension is internal, so it cancels. The only external net force along the direction of motion is the weight of the hanging block.
3. Calculate total mass and net force:
4. $$M_{\text{total}} = 3.0 + 1.5 = 4.5 \text{ kg}, \quad F_{\text{net}} = m_{\text{hanging}} g = 1.5 (9.8) = 14.7 \text{ N}$$
5. Apply Newton's second law to the system:
6. $$a = \frac{F_{\text{net}}}{M_{\text{total}}} = \frac{14.7}{4.5} \approx 3.3 \text{ m/s}^2$$
7. To find tension after solving for acceleration, isolate the table block: $T = m_{\text{table}} a = 3.0 (3.3) = 9.9 \text{ N}$.

**Worked example:** A 1500 kg elevator accelerates upward uniformly from rest to 4.0 m/s in 8.0 s. What is the tension in the elevator cable during acceleration, compared to when stationary?

1. First find acceleration from kinematics:
2. $$a = \frac{\Delta v}{\Delta t} = \frac{4.0 - 0}{8.0} = 0.50 \text{ m/s}^2 \text{ upward}$$
3. Choose upward as positive, apply Newton's second law:
4. $$F_{\text{net}} = T - mg = ma$$
5. Rearrange for tension:
6. $$T = m(g + a) = 1500(9.8 + 0.50) = 15450 \text{ N}$$
7. When stationary, $a=0$, so tension is $T_0 = mg = 14700 \text{ N}$. Tension during upward acceleration is 5% larger than stationary tension.

## Common pitfalls

- **Wrong:** Writing $a = v/F_{net}$ instead of $a = F_{net}/m$, or rearranging incorrectly for unknown variables.
  - Why it fails: Confuses proportionality relationships by memorizing rearranged forms instead of starting from the original law.
  - Correct: Always write $\vec{F}_{net} = m\vec{a}$ at the top of your solution, then rearrange step-by-step for whatever variable you need.
- **Wrong:** Decomposing weight on an incline as $W_{\parallel} = mg \cos\theta$ and $W_{\perp} = mg \sin\theta$.
  - Why it fails: Swaps sine and cosine by mixing up the angle in the weight decomposition right triangle.
  - Correct: Always check with the $\theta = 0^\circ$ (flat ground) edge case: parallel acceleration should be zero, so $\sin 0^\circ = 0$ confirms sine belongs to the parallel component.
- **Wrong:** Using weight directly (in Newtons) in place of mass in $F_{net} = ma$ when a problem gives an object's weight instead of mass.
  - Why it fails: Confuses mass (inertia, kg) and weight (force, N), even when the problem explicitly gives weight.
  - Correct: If a problem states a "50 N crate", first calculate mass from $m = W/g$ before plugging into Newton's second law.
- **Wrong:** Including tension between connected objects when calculating net force for the full system.
  - Why it fails: Forgets that internal action-reaction pairs cancel out, so they do not contribute to the system's acceleration.
  - Correct: Before calculating net force for a system, cross out any force that acts between two objects both inside your system boundary.
- **Wrong:** Assigns positive signs to both the normal force and the perpendicular component of weight on an incline, leading to non-zero acceleration perpendicular to the surface.
  - Why it fails: Poor sign convention aligned to the coordinate system.
  - Correct: Always draw your coordinate system on the free-body diagram, then assign a sign to every force based on whether it points along the positive or negative axis before summing.
- **Wrong:** Setting net force perpendicular to a fixed incline equal to $ma$.
  - Why it fails: Applies Newton's second law uniformly without accounting for the physical constraint of the incline.
  - Correct: For any fixed incline, explicitly note that $a_{\perp} = 0$, so net force perpendicular to the incline is zero.

## Cheatsheet

| Category | Formula | Notes |
| --- | --- | --- |
| Newton's Second Law (Vector) | $\vec{F}_{\text{net}} = m\vec{a}$ | Applies to any rigid object; holds for each vector component independently |
| 1-D Component Form | $F_{\text{net},x} = ma_x; \quad F_{\text{net},y} = ma_y$ | Use after resolving all forces into coordinate axis components |
| Weight | $W = mg$ | Gravitational force, units of Newtons; always acts downward near Earth's surface |
| Incline Weight (Parallel) | $W_{\parallel} = mg \sin\theta$ | Direction: down the incline; $\theta$ = incline angle from horizontal |
| Incline Weight (Perpendicular) | $W_{\perp} = mg \cos\theta$ | Direction: into the incline surface |
| Kinetic Friction | $f_k = \mu_k N$ | Direction always opposite motion relative to the surface; $N$ = normal force |
| Newton's Second Law (Systems) | $F_{\text{net,ext}} = M_{\text{total}} a$ | Internal forces cancel; only valid when all objects have same acceleration magnitude |
| Elevator Cable Tension | $T = m(g + a)$ (upward $a$) | Tension > weight for upward acceleration; $T = m(g-a)$ for downward acceleration |

## What's next

Newton's second law is the foundational quantitative tool for all of classical mechanics, so mastering it is required for every topic that comes after it in the AP Physics 1 syllabus. Next, you will apply Newton's second law to circular motion, where we relate centripetal net force to centripetal acceleration to solve problems involving orbits, banked curves, and roller coasters. Without a solid understanding of how to calculate net force and connect it to acceleration, you will not be able to correctly analyze circular motion or later topics like energy conservation and momentum, which also build on force concepts. It also connects directly to the study of simple harmonic motion, where we relate net restoring force to acceleration of oscillating objects.

- [Newton's Third Law and Free-Body Diagrams](https://www.owlsprep.com/study/ap-physics-1-u2-newton-s-third-law-and/)
- [Friction and Tension](https://www.owlsprep.com/study/ap-physics-1-u2-friction-and-tension/)
- [Inclined Planes and Atwood Machines](https://www.owlsprep.com/study/ap-physics-1-u2-inclined-planes-and-atwood-machines/)

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